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Homework Statement
What should I choose for the resistor [itex]R_{F}[/itex] and capacitor [itex]C_{F}[/itex] in this circuit?
I have an unmounted silicon photodiode, with specs:
Reverse Voltage = 100 [itex]V[/itex]
Short Circuit Current = 50 [itex]\mu A[/itex] (typical)
Dark Current = 0.5 [itex]n A[/itex] (typical), 1.0 [itex]n A[/itex] (max)
Shunt Resistance = 250 [itex]M \Omega[/itex] (min), 500 [itex]M \Omega[/itex] (typical)
Junction Capacitance = 8 [itex]pF[/itex] (typical)
Breakdown Voltage = 75 [itex]V[/itex] (typical)
I'm trying to figure out how to properly use it. Laser light from an ~800nm diode laser will be incident on it, and I would like to view its output signal on an oscilloscope via a BNC coaxial cable.
I've learned that I can operate the photodiode in "photovoltaic mode", where there is zero bias and low dark current, but slow response time, or "photoconductive mode", where there is a negative bias and fast response time, but higher dark current.
Due to the circumstances of my experiment, I'm choosing to go with "photoconductive mode". Hence, the above circuit has a negative bias. I got it from AP Technologies website, where I also saw some useful formulas:
Homework Equations
[tex]V_{OUT} = I_{S} R_{F}[/tex]
where [itex]I_{S}[/itex] is the photodiode light signal current and [itex]R_{F}[/itex] is the feedback resistance.
[tex]C_{F} = \frac{1}{2} \pi f R_{F}[/tex]
where [itex]C_{F}[/itex] is the feedback capacitance and [itex]f[/itex] is the maximum operating frequency.
The Attempt at a Solution
I guess I'm just not sure what to use for values for calculating the resistance and capacitance needed. I've searched around online, and even though I've found some useful documents on this, I'm still left with two questions:
1. How large of a resistor do I need to not fry the diode?
2. What capacitor should I use? What is the "maximum operational frequency" referring to? Is it related to the RC time constant, or to the frequency of the incident light, etc?
My naive guess for the first one would be to use [itex]V_{OUT} \sim 10 V[/itex] and [itex]I_{S} \sim 10 \mu A[/itex] so that
[tex]R_{F} = \frac{10 V}{10 \mu A} = 1 M \Omega[/tex]
That's big, though!